12/26/2015rd1 Engineering Economic Analysis Chapter 4  More Interest Formulas

12/26/2015rd2 Annual Percentage Rate (APR) or r Nominal rate 6% per year is designated ~ i. Effective interest rate ~ i eff i eff = (1 + r/m) m – 1 where m is the number of pay periods APR is 12% compounded monthly i eff = ( /12) 12 – 1 = 12.68% effective yearly rate. APR is 12% compounded monthly; find effective quarterly rate. i eff = ( /03) 3 – 1 = 3.03% effective quarterly rate. Effective Interest Rate

12/26/2015rd3 Effective Interest Rate Annual Percentage Rate (APR) is 12% If compounded monthly, effective monthly rate is 1% effective quarterly rate is 3.03% effective yearly rate is 12.68% If compounded quarterly, effective quarterly rate is 3% effective yearly rate is 12.55%

12/26/2015rd4 Interest Rate A credit card company charges 1.5% interest on the unpaid balance each month. Nominal annual interest rate is _________________. ans. 12 * 1.5% = 18% Effective annual interest rate is ________. ans. ( /12) 12 – 1 = 19.56%

12/26/2015rd5 Equivalence Are equivalent cash flows equivalent at any common point in time? For example, $1000 now at i = 10% for 10 years is equivalent to 1000( ) 10 = $ Are these 2 cash flows equivalent at time 3.37? Does $1000(1.1) 3.37 = $ (1.1) ? Check: $ = $ => Yes Are these cash flows equivalent at i = 8%? No, 1000(1.08) 3.37 = $  (1.08) = $

12/26/2015rd6 Monthly Payments a)What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly? b)Find the amount of the principal reduction of the 25 th payment. c)After making the 50 th monthly payment, you decide to pay off the loan what a check for _________. d)Find the interest on the 35 th payment. a) A = $35,000(A/P, ½ %, 60) = $ b)PR 25 = (P/F, ½ %, 60 – ) = $ c)B 50 = (P/A, ½ %, 10) = $ d)I 35 = B 34 * i = (P/A, ½ %, 26) * ½ % = $82.29

12/26/2015rd7 Mortgage Find the total interest paid on a $300, year loan at 6% compounded monthly. A = 300K(A/P, ½%, 360) = $ Total interest = 360 * $ $300K = $347,514

12/26/2015rd8 Arithmetic Gradient n -1 n Gradient begins in Year 2 = A(P/A, i%, n) + G(P/G, i%, n) G 2G (n-1)G P A

12/26/2015rd9 Gradient Example What is the present worth at pay period 0 of the following yearly cash flow at 7% compounded annually: n cf PW(7%) = (P/A, 7%, 5) + 300(P/G, 7%, 5) = = $ (F/P (PGG ) 7 1)  $

12/26/2015rd10 Geometric Gradient $100 grows geometrically by 10% per year. Compute the growth after n years n Cash Flow * 100 = 100( ) 1 = * 110 = 100( ) 2 = * 121 = 100( ) 3 = 133 ………………… n 100( ) n – 1 A n = A 1 (1 + g) n-1 Each A n must be brought back to year 0 to find the present worth.

12/26/2015rd11 Geometric Gradient A n = A n-1 (1+ g) => A n = A 1 (1 + g) n – 1 Then P =  A 1 (1 + g) n-1 (1 + i) -n P = =

12/26/2015rd12 Geometric Gradient Example A machine’s first year cost is $1,000 and increases 8% per year thereafter for 15 years. Maintenance funds earn 10% per year compounded annually. How much should be deposited in the maintenance fund to cover costs? P = 1000[1 – (1.08) 15 ( ) -15 ]/ (0.10 – 0.08) = $12, (PGGG ) 

12/26/2015rd13 Problem Find the present worth of a cash flow beginning at $10K and increasing at 8% for 4 years at 6%/yr interest. (PGGG-table ) n Cash-flow 8% PW-factor 6% PWorth $38, PW = 10K[(1 – (1.08) 4 )/(1.06) 4 (0.06 – 0.08)] = $38, (PGGG 10E ) 

12/26/2015rd14 Geometric Gradient Example You want to accumulate $1M 20 years from now by depositing $A 1 at year 1 and increasing the deposit by 6% each year for 20 years. Find A 1 if the bank pays 8% interest compounded annually. F = A 1 (P/A1, g = 6%, i = 8%, n = 20)(F/P, i = 8%, 20) 1M = A1{[1 – * ]/0.02}(1.08) 20 A1 = $13, = (/ 1e6 (FGP (PGGG ) 8 20) P =

12/26/2015rd15 Effective Interest Rates $1000 is deposited at 7% compounded monthly. Find the value 5 years from now using monthly, quarterly, semiannually, yearly and bi-yearly effective rates. Monthly: 1000(F/P, 7/12%, 60) = $ Quarterly: 1000(F/P, 1.76%, 20) = $ Semi-annually 1000(F/P, 3.55%, 10) = $ Annually 1000(F/P, 7.23%, 5) = $ Biennially 1000(F/P, 14.98%, 2.5) = $ Pentad 1000(F/P, 41.76%, 1) = $ e.g. Pentad effective rate = [ /60) 60 – 1 = % 1000(1 + i m ) nm = 1000(1 + i a ) n

12/26/2015rd16 Relationships of Interest Factors F/P =1/(P/F); A/P = 1/(P/A); F/A = 1(A/F) F/A = 1 + Σ(F/P, i%, n-1) A/P = A/F + i; A/P = P/A – A/F; CRF = (P-S)(A/P. i%, n) + Si P/F * F/A = P/A; P/F = 1 – P/A * i A/F = A/P – i; 7% n F/P P/F A/F A/P F/A P/A A/G P/G

12/26/2015rd17 Continuous Compounding In the effective interest formula let m = rp and the formula becomes i eff = (1 + r/m) m - 1 e i eff = (1 + 1/p) rp = = e r as p   (F/P, r%, N) = e rN for continuous compounding

12/26/2015rd18 Continuous Compounding You deposit $100 per month in a savings account with an APR of 6% per year compounded continuously. How much will accumulate in 5 years? F = 100(F/A, e , 60) = $ The monthly continuous compounding rate e = 0.501

12/26/2015rd19 Problem 4-36 $12K is borrowed at 4% per annum and is to repaid in 5 payments. After the 2 nd payment, the borrower was given the option of paying off the loan the following year. How much was then due? A = 12K(A/P, 4%, 5) = $ Balance = (P/A, 4%, 2) = $

12/26/2015rd20 Problem 4-38 Sold in 2002 for $150K at 20% down payment and 15-year loan at 8% per year. Buyer makes first payment in How much will be owed after 2009? Loan amount = 150K * 150K = $120K A = 120K(A/P, 8% 15) = $14, Balance after making 7 payments is B = 14, (P/A, 8%, 8) = $80, (loan 120E3 8 15)

12/26/2015rd21 Problem 4-50 A debt of $5K is repaid according to the cash flow below at 8% compound interest. Find X. n cf $ X [5K – [500(P/A, 8%, 4) + 500(P/G 8% 4)](F/P, 8% 5) = X => X = $ (List-pgf '( ) 8)  $5000 (IRR '( ))  8%

12/26/2015rd22 Capitalized Cost You can have 5% interest in perpetuity (forever). You need to generate $10,000 a year for a scholarship fund. How much investment is needed to do so? P = A/i = 10,000/0.05 = $200,000.

12/26/2015rd23 Gradient Find the equivalent sum at year 7 for the following cash flow at 7% compound interest per year. n Cf F 7 = [(1000(P/A, 7%, 5) (P/G, 7%, 5) + 50(P/F 7%, 4)](F/P, 7%, 7) = $18, (F/P (+ (PGG ) (P/F )) 7 7) 

12/26/2015rd24 Shady Deal You borrow $1000 to be repaid in 24 monthly installments. The interest rate is a mere 1.5% per month. Further Amount requested$1000 Credit risk insurance 5 Credit investigation 25 Total $1030 Interest: ($1030)(24)(0.015) = $371 Total owed: $ $371 = $1401 Payment: $1401/24 = $58.50 Find the effective annual interest rate charged = 58.50(P/A, i%, 24) => i m = 2.92% => APR = 34.04% I aeff = 41.25%

12/26/2015rd25 Sports Contracts Headline blares Ace Stacey sings 10-year contract for $50 million paid $5M now and $4M for the first 5 years and $5M for the next 5 years. How much is the contract worth to Ace now if the interest rate is 7%? PW = 5M + 4M(P/A, 7%, 5) + 5M(P/A, 7%, 5)(P/F, 7%, 5) = 5M + 16,400, ,616,921 = $36,017,710

12/26/2015rd26 Capital Recovery (A/P, i%, n) – (A/F, i%, n) = i P F = Salvage P(A/P, i%, n) – S(A/F, i%, n) EUAC = (P - S)(A/P, i%, n) + Si

12/26/2015rd27 Capital Recovery Example A new machine's first cost is $5,000 with a 5-year life and a salvage value of $1000. Compute the annual cost at i = 7%. 1)5000(A/P, 7%, 5) – 1000(A/F, 7%, 5) = – = $ )(P – S)(A/P, 7%, 5) + Si = 4000(P/A, 7%,5) *1000 = = $ ) (P – F)(A/F, 7%, 5) + Pi = 4000(A/F, 7%, 5) * 0.07 = = $

Review 12/26/2015rd28

12/26/2015rd29 Change in Rate You borrow $20,000 at 7% compounded monthly over 48 months. After making the 24 th payment, you negotiate with the bank to pay off the loan in 8 equal quarterly payments. Determine the quarterly payment at the same interest rate. A m = 20K(A/P, 7/12 %, 48) = $ B 24 = (P/A, 7/12 %, 24) = $10, A q = 10,698.84(A/P, 1.76%, 8) = $

Exact Rate of Return Find the exact rate of return for the following cash flow. n cf (quadratic )  => % (list-pgf '( ) )  0 (IRR '( ))  /26/2015rd30

Wright Learning Curve Unit Hours a) The time to make the 10 th unit is ________. b)The learning curve rate in percent is ________. c)The slope of the learning curve is ________. d)The time to make the 13th unit is __________. 12/26/2015rd31

Mortgage You borrow $10,000 at 6% compounded monthly for 24 years. Your monthly payment is closest to a) $660b) $550c) $450d) $350 Your principal reduction on the 12 th payment is closest to a) $315b) $415c) $515 d) not given Total interest paid after making 12 th payment is a) $370b) $470c) $570d) $670 12/26/2015rd32

Annual Worth You want $100,000 in a fund 10 years from now, the amount to deposit in years 6 through 9 at i = 10% per year is closest to a) $19,588b) $20,614c) $21,547d) $22,389 $100K (AGF (PGF 100E3 10 1) 10 4)  /26/2015rd33

Doubling Investment If you invest $2,000 at 12% compounded monthly for the same length of time that it takes an investment to double in value at 12% compounded quarterly, you will have a) $3709b) $4027c) $4352d) = (1.03) q => q = quarters or months 2K(1.01) = $ /26/2015rd34

12/26/2015rd35 Measuring Investments 1.Present Worth (PW) 2.Annual Worth (AW) 3.Future Worth 4.Internal Rate of Return 5.External Rate of Return 6.Benefits/Costs ratio 7.Payback Period 8.Capitalized Worth